Calculate Volume using the Method of Cylindrical Shells - Urgent

**LightRay7** · October 7th 2012, 04:58 AM

Here are some things I have done. I already found the volume since the other part of the question said to use the method of discs/washers. It is 54.0633 cubic units (correct answer).

I found by trial and error that (since this is an online assignment) that a= 9/sqrt(82) and c =9 .

What do I do? Why should there be two integrals, I don't understand.

I have tried the following.

y = 9^x becomes x = lny/ln9

Formula

h = lny/ln9
r = y

So I got integral from 9/sqrt(82) to 9 of

Calculate Volume using the Method of Cylindrical Shells - Urgent-capture.png

dy

But that's just one integral and perhaps even incorrect. Please show me what to do.

Thanks

**skeeter** · October 7th 2012, 05:58 AM

the sketch of the graph is misleading ...

note that $\frac{9}{\sqrt{82}} < 1$ , so there is a very small region between the curve $y = \frac{9}{\sqrt{x^2+81}}$ and $x=1$ from $y = \frac{9}{\sqrt{82}}$ to $y = 1$ .

**Prove It** · October 7th 2012, 05:59 AM

To evaluate the volume of your region, you need to do

$\displaystyle \begin{align*} \int_1^9{ \pi \cdot 1^2 \, dy } - \int_1^9{ \pi \cdot \left( \log_9{y} \right)^2 \,dy } + \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot 1^2 \, dy } - \int_{ \frac{9}{ \sqrt{82} } }^1{ \pi \cdot \left[ \sqrt{ \left( \frac{y}{9} \right)^{ -2 } - 81 } \right]^2 \,dy } &= \pi \int_1^9{ 1 \, dy } - \frac{\pi}{\left( \ln{9} \right)^2} \int_1^9{ \left( \ln{y} \right)^2 \,dy } + \pi \int_{ \frac{9}{\sqrt{82}} }^1{ 1\, dy } - 81\pi \int_{\frac{9}{\sqrt{82}}}^1{ y^{-2} - 1 \, dy } \end{align*}$

The only one of those integrals which may cause you any trouble is the second, so do

$\displaystyle \begin{align*} \int{\left( \ln{y} \right)^2 \, dy} &= \int{ \frac{y\left( \ln{y} \right)^2}{y} \, dy } \\ &= \int{ u^2 \, e^u \, du } \textrm{ after making the substitution } u = \ln{y} \end{align*}$

which you would then evaluate using integration by parts twice. See how you go

**LightRay7** · October 7th 2012, 12:02 PM

Is this the method of cylindrical shells? I am obliged to use cylindrical shells. Can you use r (radius), h(height)? And why will my answer be in the form of a sum of two integrals (see the question -top of image)?

**skeeter** · October 7th 2012, 12:58 PM

using the method of cylindrical shells ...

$V = 2\pi \int_{\frac{9}{\sqrt{82}}}^1 y \left(1 - \frac{9\sqrt{1-y^2}}{y} \right) \, dy + 2\pi \int_1^9 y \left(1 - \frac{\ln{y}}{\ln{9}}\right) \, dy$

**LightRay7** · October 7th 2012, 03:51 PM

It worked.

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